% !TEX program = xelatex
\documentclass[12pt]{article}
\usepackage[text width=16cm,text height=23cm]{geometry}
\usepackage{amsmath}
\usepackage{gensymb}
\usepackage{amssymb}
\usepackage{ctex}
\usepackage{tkz-euclide}


\newcommand{\pingxing}{/\hspace{-0.2em}/}
\newcommand{\zheng}{{\bf 证：}}
\newcommand{\zhengming}{{\bf 证明：}}
\newcommand{\jie}{{\bf 解：}}

\begin{document}
\section*{第一题：}

\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：$\triangle ABC$外接于$\odot O$，$\angle BAC=60\degree$，$AE\perp BC$，$CF\perp AB$，$AE$、$CF$相交于点$H$，点$D$为弧$BC$的中点，连接$HD$、$AD$。求证：$\triangle ABC$为等腰三角形。

	\noindent\zheng

	\vspace{5em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		%绘制三角形ABC
		\tkzDefPoints{0/0/B,4/0/C}
		\tkzDefPoint(70:3.5){A}
		\tkzDrawPolygon(A,B,C)
		%作BC和AB边上的高
		\tkzDefPointBy[projection=onto B--C](A)
		\tkzGetPoint{E}
		\tkzDefPointBy[projection=onto A--B](C)
		\tkzGetPoint{F}
		\tkzDrawSegments(A,E C,F)
		%画AE和CF的交点H，三角形的高的交点
		\tkzDefTriangleCenter[ortho](B,C,A)
		%如果是两条直线的交点，则用下面的命令
		%\tkzInterLL(A,E)(C,F)
		\tkzGetPoint{H}
		%作三角形ABC的外接圆
		\tkzCircumCenter(A,B,C)\tkzGetPoint{O}
		\tkzDrawCircle(O,A)
		%连接OA，OB，OC
		\tkzDrawSegments[dashed](O,A O,B O,C)
		%画点D，并连接BD，CD，OD，BH
		%方法一：直接定义点D
		\tkzDefPoints{2/-1.17/D}
		%方法二：作O点关于直线AB的对称点，由于作图不是很标准，D点没有落到圆上
		%\tkzDefPointBy[reflection = over B--C](O)
		%\tkzGetPoint{D}
		\tkzDrawSegments[dashed](O,D B,D C,D B,H)
		\tkzDrawSegments(D,H)
		%画出所有点
		\tkzDrawPoints[size=1.5pt](A,B,C,D,E,F,H,O)
		%标注点
		\tkzLabelPoints[above left](A,F,H)
		\tkzLabelPoints[below left](B)
		\tkzLabelPoints[below right](C)
		\tkzLabelPoints[above right](O)
		\tkzLabelPoints[below](D,E)
	\end{tikzpicture}
\end{minipage}




\section*{第二题：}

\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 如图，$F$为正方形$ABCD$边$CD$上一点，连接$AC$、$AF$，延长$AF$交$AC$的平行线$DE$于点$E$，连接$CE$，且$AC=AE$。求证：$CE=CF$。

	\noindent\zheng

	\vspace{6em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=.8]
		%绘制正方形ABCD
		\tkzDefPoints{0/0/B,4/0/C}
		\tkzDefSquare(B,C)\tkzGetPoints{D}{A}
		\tkzDrawSquare(B,C)
		%画F点
		\tkzDefPoints{4/3/F}
		%延长AF到E，使得AC＝AE
		\tkzDefBarycentricPoint(A=0.372,F=-1.372)
		\tkzGetPoint{E}
		%作E点关于AD的对称点
		\tkzDefPointBy[reflection=over A--D](E)
		\tkzGetPoint{G}
		%连线
		\tkzDrawSegments(A,C D,E C,E A,E)
		\tkzDrawSegments[dashed](A,G D,G C,G)
		%画点
		\tkzDrawPoints(A,B,C,D,E,F,G)
		%标点
		\tkzLabelPoints[left=0](A,B)
		\tkzLabelPoints[below=0](C)
		\tkzLabelPoints[above left=0](D)
		\tkzLabelPoints[right=0](E)
		\tkzLabelPoints[above right=0](F,G)
	\end{tikzpicture}
\end{minipage}

\section*{第三题：}

\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：$\triangle ABC$中，$AB=AC$，$\angle BAC=20\degree$，$\angle BDC=30\degree$，求证：$AD=BC$

	\noindent\zheng

	\vspace{10em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		%绘制三角形
		\tkzDefPoints{0/0/B,2/0/C}
		\tkzDefPoint(80:6){A}
		\tkzDrawPolygon(A,B,C)
		%画D点
		\tkzDefPoint(80:3.76){D}
		%画等边三角形ADE
		\tkzDefTriangle[equilateral](D,A)\tkzGetPoint{E}
		%画点
		\tkzDrawPoints(A,B,C,D,E)
		%连线
		\tkzDrawSegments(C,D)
		\tkzDrawSegments[dashed](A,E D,E C,E)
		%标点
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[below left=0](B)
		\tkzLabelPoints[below right=0](C)
		\tkzLabelPoints[right=0](D)
		\tkzLabelPoints[left=0](E)
	\end{tikzpicture}
\end{minipage}

\section*{第四题：}

\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：$\triangle ABC$中，$D$为$AC$边的中点，$\angle A=3\angle C$，$\angle ADB=45\degree$。求证：$AB\perp BC$。

	\noindent\zheng

	\vspace{5em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=1.3]
		%绘制三角形
		\tkzDefPoints{0/0/A,4/0/C}
		\tkzDefPoint(67.5:1.53){B}
		\tkzDrawPolygon(A,B,C)
		%找AC中点
		\tkzDefMidPoint(A,C)\tkzGetPoint{D}
		%找点E,方法是作AC的中垂线，取中垂线上一点e，得到BC与De的交点就是E
		\tkzDefLine[mediator](A,C)\tkzGetPoint{e}
		\tkzInterLL(B,C)(D,e)\tkzGetPoint{E}
		\tkzDrawPoints(A,B,C,D,E)
		\tkzDrawSegments[dashed](A,E D,E)
		\tkzDrawSegments(B,D)
		\tkzLabelPoints[below=0](A,C,D)
		\tkzLabelPoints[above=0](E)
		\tkzLabelPoints[above left=0](B)
	\end{tikzpicture}
\end{minipage}

\section*{第五题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 如图，四边形$ABCD$的两条对角线$AC$，$BD$交于点$E$，$\angle BAC=50\degree$，$\angle ABD=60\degree$，$\angle CBD=20\degree$，$\angle CAD=30\degree$，$\angle ADB=40\degree$。求$\angle ACD$。

	\noindent\jie

	\vspace{12em}
\end{minipage}
\hspace{1em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=.8]
		\tkzDefPoints{0/0/B,4/0/D}
		\tkzDefPoint(60:2.61){A}
		\tkzDefPoint(-20:2.61){C}
		\tkzDrawPolygon(A,B,C,D)
		\tkzDrawSegments(A,C B,D)
		\tkzInterLL(A,C)(B,D)\tkzGetPoint{E}
		%通过绘制已知两个角的三角形ABF得到F点
		\tkzDefTriangle[two angles=80 and 80](A,B)\tkzGetPoint{F}
		%过D点作AB的平行线
		\tkzDefLine[parallel=through D](A,B)\tkzGetPoint{g}
		\tkzInterLL(B,F)(D,g)\tkzGetPoint{G}
		\tkzInterLL(A,G)(B,D)\tkzGetPoint{H}
		\tkzDrawPoints(A,B,C,D,E,F,G,H)
		\tkzDrawSegments[dashed](A,F B,F A,G D,G C,H)
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[left=0](B)
		\tkzLabelPoints[below left=-3pt](C,E,G)
		\tkzLabelPoints[above right=-3pt](D,H)
		\tkzLabelPoints[right=0](F)
	\end{tikzpicture}
\end{minipage}

\section*{第六题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知，$\angle ABC=30\degree$，$\angle ADC=60\degree$，$AD=DC$.求证：$AB^2+BC^2=BD^2$.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/D,5.58/0/B}
		\tkzDefPoint(18:2){A}
		%定义等边三角形ADC
		\tkzDefTriangle[equilateral](A,D)\tkzGetPoint{C}
		\tkzDrawPolygon(A,D,C,B)
		\tkzDrawSegments(D,B)
		\tkzDefTriangle[equilateral](A,B)\tkzGetPoint{E}
		\tkzDrawSegments[dashed](A,C A,E B,E C,E)
		\tkzDrawPoints(A,B,C,D,E)
		\tkzLabelPoints[above left=0](A)
		\tkzLabelPoints[right=0](B)
		\tkzLabelPoints[below=0](C)
		\tkzLabelPoints[left=0](D)
		\tkzLabelPoints[above=0](E)
	\end{tikzpicture}
\end{minipage}

\section*{第七题：}
\noindent 如图，$PC$切$\odot O$于$C$，$AC$为圆的直径，$PEF$为$\odot O$的割线，$AE$、$AF$与直线$PO$相交于$B$、$D$.求证：四边形$ABCD$为平行四边形.

\vspace{1em}\noindent
\begin{minipage}{0.4\textwidth}
	\noindent\zhengming

	\vspace{15em}
\end{minipage}
\hspace{0.5em}
\begin{minipage}{0.5\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/O,-6.7/0/P,1.6/0/D}
		\tkzDefPoint(68:2.5){A}
		%以OA为半径定义圆，O为圆心，并画出圆O
		\tkzDefCircle[through](O,A)\tkzDrawCircle(O,A)
		%定义圆O过点P的切线
		\tkzDefTangent[from=P](O,A)\tkzGetPoints{C}{c}
		%用直线AD和圆的交点找点F
		\tkzInterLC(A,D)(O,A)\tkzGetPoints{f}{F}
		%找PF和圆的另一个交点E
		\tkzInterLC(P,F)(O,A)\tkzGetPoints{}{E}
		%AE和PD的交点B
		\tkzInterLL(A,E)(P,D)\tkzGetPoint{B}
		%过C作PD的垂线交PD于点G
		\tkzDefPointBy[projection=onto P--D](C)\tkzGetPoint{G}
		\tkzDrawPoints(A,B,C,D,E,F,G,O,P)
		\tkzDrawSegments(P,D P,F P,C A,E A,F A,C B,C)
		\tkzDrawSegments[dashed](C,E C,G C,F G,E G,F O,E O,F)
		\tkzLabelPoints[above right=0pt](A)
		\tkzLabelPoints[above left=0](B)
		\tkzLabelPoints[below left=0pt](C,E)
		\tkzLabelPoints[right=0](D)
		\tkzLabelPoints[below right=0](F)
		\tkzLabelPoints[above](P,G)
		\tkzLabelPoints[above right=-3pt](O)
	\end{tikzpicture}
\end{minipage}



\section*{第八题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：在$\triangle ABC$中，$AB=AC$，$\angle A=80\degree$，$\angle OBC=10\degree$，$\angle OCA=20\degree$.求证：$AB=OB$.

	\noindent\zheng

	\vspace{5em}
\end{minipage}
\hspace{1em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=.85]
		\tkzDefPoints{0/0/B}
		\tkzDefPoint(50:5){A}
		\tkzDefTriangle[two angles=80 and 50](A,B)\tkzGetPoint{C}
		\tkzDrawPolygon(A,B,C)
		\tkzDefPoint(10:5){O}
		\tkzDrawSegments(O,A O,B O,C)
		\tkzInterLL(C,O)(A,B)\tkzGetPoint{D}
		\tkzDefTriangle[equilateral](C,O)\tkzGetPoint{E}
		\tkzDrawSegments[dashed](O,D O,E A,E C,E)
		\tkzDrawPoints(A,B,C,D,E,O)
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[left=0](B)
		\tkzLabelPoints[right=0](C)
		\tkzLabelPoints[above left=-5pt](D)
		\tkzLabelPoints[below=0](E)
		\tkzLabelPoints[above right=-5pt](O)
	\end{tikzpicture}
\end{minipage}

\section*{第九题：}
\noindent
\begin{minipage}{0.65\textwidth}
	\noindent 已知：正方形$ABCD$中，$\angle OAD=\angle ODA=15\degree$，求证：$\triangle OBC$为正三角形.

	\noindent\zheng

	\vspace{5em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.25\textwidth}
	\begin{tikzpicture}[scale=.8]
		\tkzDefPoints{0/0/B,4/0/C}
		\tkzDefTriangle[equilateral](B,C)\tkzGetPoint{O}
		\tkzDrawPolygon(B,C,O)
		\tkzDefSquare(B,C)\tkzGetPoints{D}{A}
		\tkzDrawSegments(A,B C,D A,D O,A O,D)
		\tkzDrawPoints(A,B,C,D,O)
		\tkzLabelPoints[left=0](A,B)
		\tkzLabelPoints[right=0](C,D)
		%\tkzLabelPoints[below left=0](O)
		\tkzLabelPoints[below=2pt,label=$O^{'}$](O)
	\end{tikzpicture}
\end{minipage}


\section*{第十题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：正方形$ABCD$中，$E$、$F$为$AD$、$BC$的中点，连接$BE$、$AF$，相交于点$P$，连接$PC$.求证：$PC=BC$.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.55\textwidth}
	\begin{tikzpicture}[scale=.8]
		\tkzDefPoints{0/0/B,5/0/C}
		\tkzDefSquare(B,C)\tkzGetPoints{D}{A}
		\tkzDrawPolygon(A,B,C,D)
		\tkzDefMidPoint(A,D)\tkzGetPoint{E}
		\tkzDefMidPoint(D,C)\tkzGetPoint{F}
		\tkzInterLL(A,F)(B,E)\tkzGetPoint{P}
		\tkzDrawSegments(A,F B,E P,C)
		\tkzDrawSegments[dashed](B,F)
		\tkzDrawPoints(A,B,C,D,E,F,P)
		\tkzLabelPoints[left=0](A,B)
		\tkzLabelPoints[right=0](C,D,F)
		\tkzLabelPoints[above=0](E,P)
	\end{tikzpicture}
\end{minipage}

\section*{第十一题：}
\noindent
\begin{minipage}{0.55\textwidth}
	如图，$\triangle ACB$与$\triangle ADE$都是等腰三角形，$\angle ADE=\angle ACB=90\degree$，$\angle CDF=45\degree$，$DF$交$BE$于$F$，求证：$\angle CFD=90\degree$.

	\noindent\zhengming

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/C,4/0/B,-0.5/1.1/E}
		\tkzDefPoint(90:4){A}
		\tkzDrawPolygon(A,B,C)
		\tkzDefMidPoint(B,E)\tkzGetPoint{F}
		\tkzDefTriangle[two angles=45 and 45](A,E)\tkzGetPoint{D}
		\tkzDrawPolygon(A,E,D)
		\tkzDrawSegments(A,E B,E C,F D,F C,D)
		\tkzDrawPoints(A,B,C,D,E,F)
		\tkzLabelPoints[left=0](A,C,D,E)
		\tkzLabelPoints[below=0](F)
		\tkzLabelPoints[right=0](B)
	\end{tikzpicture}
\end{minipage}

\section*{第十二题：}
\noindent
\begin{minipage}{0.55\textwidth}
	已知：在$\triangle ABC$中，$\angle CBA=2\angle CAB$，$\angle CBA$的角平分线$BD$与$\angle CAB$的角平分线$AD$相交于点$D$，且$BC=AD$.求证：$\angle ACB=60\degree$.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/B,5/0/A}
		\tkzDefTriangle[two angles=80 and 40](B,A)\tkzGetPoint{C}
		\tkzDrawPolygon(A,B,C)
		\tkzDefLine[bisector](C,B,A)\tkzGetPoint{b}
		\tkzDefLine[bisector](B,A,C)\tkzGetPoint{a}
		\tkzInterLL(B,b)(A,a)\tkzGetPoint{D}
		\tkzDefLine[bisector](A,B,D)\tkzGetPoint{e}
		\tkzInterLL(B,e)(A,C)\tkzGetPoint{E}
		\tkzDrawSegments(A,D B,D)
		\tkzDrawSegments[dashed](B,E D,E)
		\tkzDrawPoints(A,B,C,D,E)
		\tkzLabelPoints[below right=-3pt](A)
		\tkzLabelPoints[below left=-3pt](B)
		\tkzLabelPoints[above=0](C,D)
		\tkzLabelPoints[above right=-3pt](E)
	\end{tikzpicture}
\end{minipage}

\section*{第十三题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：在$\triangle ABC$中，$AC=BC$，$\angle C=100\degree$，$AD$平分$\angle CAB$.求证：$AD+CD=AB$.

	\noindent\zheng
	\vspace{9em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/A,5/0/B}
		\tkzDefTriangle[two angles=40 and 40](A,B)\tkzGetPoint{C}
		\tkzDrawPolygon(A,B,C)
		\tkzDefLine[bisector](C,A,B)\tkzGetPoint{a}
		\tkzInterLL(A,a)(B,C)\tkzGetPoint{D}
		\tkzDefTriangle[two angles=40 and 80](A,B)\tkzGetPoint{E}
		\tkzInterLL(A,D)(B,E)\tkzGetPoint{F}
		\tkzDrawSegments(A,D)
		\tkzDrawPolygon[dashed](C,E,F)
		\tkzDrawSegments[dashed](B,F D,E D,F)
		\tkzDrawPoints(A,B,C,D,E,F)
		\tkzLabelPoints[above left=-3pt](A,C)
		\tkzLabelPoints[right=0](B,F)
		\tkzLabelPoints[below=0](D)
		\tkzLabelPoints[above=0](E)
	\end{tikzpicture}
\end{minipage}

\section*{第十四题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：在$\triangle ABC$中，$AB=BC$，$D$是$AC$的中点，过$D$作$DE\perp BC$于$E$，连接$AE$，取$DE$中点$F$，连接$BF$.求证：$AE\perp BF$.

	\noindent\zheng
	\vspace{9em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/A,6/0/C}
		\tkzDefTriangle[two angles=37 and 37](A,C)\tkzGetPoint{B}
		\tkzDrawPolygon(A,B,C)
		\tkzDefMidPoint(A,C)\tkzGetPoint{D}
		\tkzDefPointBy[projection=onto B--C](D)\tkzGetPoint{E}
		\tkzDefMidPoint(D,E)\tkzGetPoint{F}
		\tkzInterLL(A,E)(B,F)\tkzGetPoint{G}
		\tkzDrawSegments(A,E B,F D,E)
		\tkzDrawSegments[dashed](B,D D,G)
		\tkzDrawPoints(A,B,C,D,E,F,G)
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[above right=-3pt](B,C,E,G)
		\tkzLabelPoints[below=0](D)
		\tkzLabelPoints[right=0](F)
	\end{tikzpicture}
\end{minipage}

\section*{第十五题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 已知：在$\triangle ABC$中，$\angle A=24\degree$，$\angle C=30\degree$，$D$为$AC$上一点，$AB=CD$，连接$BD$.求证：$AB\cdot BC=BD\cdot AC$.

	\noindent\zheng
	\vspace{9em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/A,6/0/C}
		\tkzDefTriangle[two angles=24 and 30](A,C)\tkzGetPoint{B}
		\tkzDrawPolygon(A,B,C)
		\tkzDefTriangle[equilateral](B,A)\tkzGetPoint{O}
		\tkzDefTriangle[two angles=72 and 36](O,C)\tkzGetPoint{D}
		\tkzDrawSegments[dashed](O,A O,B O,C O,D)
		\tkzDrawSegments(B,D)
		\tkzDrawPoints(A,B,C,D,O)
		\tkzLabelPoints[left=0](A)
		\tkzLabelPoints[above=0](B)
		\tkzLabelPoints[right=0](C)
		\tkzLabelPoints[below left=-3pt](D)
		\tkzLabelPoints[below=0](O)
	\end{tikzpicture}
\end{minipage}

\section*{第十六题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 如图，在$\triangle ABC$三边上，向外做三角形$ABR$、$BCP$、$CAQ$，使$\angle CBP=\angle CAQ=45\degree$，$\angle BCP=\angle ACQ=30\degree$，$\angle ABR=\angle BAR=15\degree$.求证：$RQ$与$RP$垂直且相等.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}
		\tkzDefPoints{0/0/A,4/0/B,0.8/5/C}
		\tkzDrawPolygon(A,B,C)
		\tkzDefTriangle[two angles=45 and 30](A,C)\tkzGetPoint{Q}
		\tkzDefTriangle[two angles=30 and 45](C,B)\tkzGetPoint{P}
		\tkzDefTriangle[two angles=15 and 15](B,A)\tkzGetPoint{R}
		\tkzDrawPolygon(A,R,B,P,C,Q)\tkzDrawSegments(Q,R R,P)
		\tkzDefTriangle[equilateral](R,B)\tkzGetPoint{O}
		\tkzDrawSegments[dashed](O,A O,B O,P O,R)
		\tkzLabelPoints[below=0](A,B,R)
		\tkzLabelPoints[above left=-3pt](C,O)
		\tkzLabelPoints[left=0](Q)
		\tkzLabelPoints[right=0](P)
	\end{tikzpicture}
\end{minipage}

\section*{第十七题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 如图，三角形$ABC$内接于$\odot O$，两条高$AD$、$BE$交于点$H$，连接$AO$、$OH$.若$AH=2$，$BD=3$，$CD=1$，求三角形$AOH$的面积.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=1.3]
		\tkzDefPoints{0/0/B,3/0/D,4/0/C}
		\tkzDefTriangle[two angles=45 and 90](B,D)\tkzGetPoint{A}
		\tkzDrawPolygon(A,B,C)
		\tkzDefPointBy[projection=onto A--C](B)\tkzGetPoint{E}
		\tkzDefMidPoint(B,C)\tkzGetPoint{F}
		\tkzDefCircle[circum](A,B,C)\tkzGetPoint{O}
		\tkzDrawCircle(O,A)
		\tkzInterLL(A,D)(B,E)\tkzGetPoint{H}
		\tkzDrawSegments(A,D A,O B,E O,H)
		\tkzDrawSegments[dashed](O,F)
		\tkzDrawPoints(A,B,C,D,E,F,H,O)
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[left=0](B,O)
		\tkzLabelPoints[right=0](C,E)
		\tkzLabelPoints[below=0](D,F)
		\tkzLabelPoints[below right=-3pt](H)
	\end{tikzpicture}
\end{minipage}

\section*{第十八题：}
\noindent
\begin{minipage}{0.55\textwidth}
	\noindent 如图，$\angle DAC=2x$，$\angle ACB=4x$，$\angle ABC=3x$，$AD=BC$，求$\angle BAD$.

	\noindent\zheng

	\vspace{8em}
\end{minipage}
\hspace{2em}
\begin{minipage}{0.35\textwidth}
	\begin{tikzpicture}[scale=.8]
		\tkzInit
		\tkzDefPoints{0/0/D,3/0/C}
		\tkzDefTriangle[two angles=72 and 72](D,C)\tkzGetPoint{A}
		\tkzDefTriangle[two angles=108 and 18](D,A)\tkzGetPoint{B}
		\tkzDefTriangle[two angles=18 and 108](A,C)\tkzGetPoint{E}
		\tkzFillAngles[size=.5cm,left color=gray!50,right color=gray!50](D,B,A D,A,C A,C,D)
		\tkzMarkAngles[size=.5cm](D,B,A D,A,C A,C,D)
		\tkzLabelAngle[pos=.8](D,B,A){\tiny{$3x$}}
		\tkzLabelAngle[pos=.8](D,A,C){\tiny{$2x$}}
		\tkzLabelAngle[pos=.8](A,C,D){\tiny{$4x$}}
		\tkzDrawPolygon(A,D,C)
		\tkzDrawSegments(A,B B,D)
		\tkzDrawSegments[dashed](A,E C,E)
		\tkzDrawPoints(A,B,C,D,E)
		\tkzLabelPoints[above=0](A)
		\tkzLabelPoints[below=0](B,C,D,E)
	\end{tikzpicture}
\end{minipage}







\end{document}